设置的最低有效位的位置

unsigned GetLowestBitPos(无符号值){断言(值！= 0)；//单独处理无符号位置 = 0;while (!(value & 1)){值＞＞＝1；++位置；}返回位置；}

a> 提供了一个很好的集合，呃，一些小技巧，并附有性能/优化讨论.对于您的问题(来自该站点)，我最喜欢的解决方案是乘法和查找":

unsigned int v;//在 32 位 v 中找到尾随零的数量国际r;//结果在这里static const int MultiplyDeBruijnBitPosition[32] ={0, 1, 28, 2, 29, 14, 24, 3, 30, 22, 20, 15, 25, 17, 4, 8,31, 27, 13, 23, 21, 19, 16, 7, 26, 12, 18, 6, 11, 5, 10, 9};r = MultiplyDeBruijnBitPosition[((uint32_t)((v & -v) * 0x077CB531U)) >>27];

有用的参考资料:

• "使用 de Bruijn 序列索引计算机单词中的 1"- 解释上述代码为何有效.
• "董事会代表 >位板 >比特扫描"- 对该问题的详细分析，特别关注国际象棋编程

I am looking for an efficient way to determine the position of the least significant bit that is set in an integer, e.g. for 0x0FF0 it would be 4.

A trivial implementation is this:

unsigned GetLowestBitPos(unsigned value)
{
   assert(value != 0); // handled separately

   unsigned pos = 0;
   while (!(value & 1))
   {
      value >>= 1;
      ++pos;
   }
   return pos;
}

Any ideas how to squeeze some cycles out of it?

(Note: this question is for people that enjoy such things, not for people to tell me xyzoptimization is evil.)

[edit] Thanks everyone for the ideas! I've learnt a few other things, too. Cool!

Bit Twiddling Hacks offers an excellent collection of, er, bit twiddling hacks, with performance/optimisation discussion attached. My favourite solution for your problem (from that site) is multiply and lookup:

unsigned int v;  // find the number of trailing zeros in 32-bit v 
int r;           // result goes here
static const int MultiplyDeBruijnBitPosition[32] = 
{
  0, 1, 28, 2, 29, 14, 24, 3, 30, 22, 20, 15, 25, 17, 4, 8, 
  31, 27, 13, 23, 21, 19, 16, 7, 26, 12, 18, 6, 11, 5, 10, 9
};
r = MultiplyDeBruijnBitPosition[((uint32_t)((v & -v) * 0x077CB531U)) >> 27];

Helpful references:

• "Using de Bruijn Sequences to Index a 1 in a Computer Word" - Explanation about why the above code works.
• "Board Representation > Bitboards > BitScan" - Detailed analysis of this problem, with a particular focus on chess programming

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